Newton’s backward interpolation
Suppose \(y=f( x )\)is tabulated for equally spaced valued
\( x=x_0,x_0+h,x_0+2h,\ldots ,x_0+nh \)
given by
\( y=y_0,y_1,y_2,\ldots ,y_n \)
Here,
\( x_i=x_0+ih \) for \( i=0,1,2,\ldots ,n \)
Now, assuming \(y=f( x )\) as a polynomial of degree n in x, we write
\( y=a_n+a_{n-1}( x-x_n )+a_{n-2}( x-x_n )( x-x_{n-1} )+\ldots +a_0( x-x_n )\ldots ( x-x_1 )\)
Putting, \( x=x_n\), we get
\( y=a_n+a_{n-1}( x-x_n )+a_{n-2}( x-x_n )( x-x_{n-1} )+\ldots +a_0( x-x_n )\ldots ( x-x_1 )\)
or \( y_n=a_n+a_{n-1}( \) \(x_n\) \(-x_n )+a_{n-2}( \) \(x_n\) \( -x_n )( \) \(x_n\) \(-x_{n-1} )+\ldots +a_0( \) \(x_n\) \( -x_n )\ldots ( \) \(x_n\) \(-x_1 )\)
or \(y_n=a_n+0+...\)
or \(y_n=a_n\)
Putting, \( x=x_{n-1}\), we get
\( y=a_n+a_{n-1}( x-x_n )+a_{n-2}( x-x_n )( x-x_{n-1} )+\ldots +a_0( x-x_n )\ldots ( x-x_1 )\)
or \( y_{n-1}=a_n+a_{n-1}( \) \(x_{n-1}\) \(-x_n )+a_{n-2}( \) \(x_{n-1}\) \( -x_n )(\) \(x_{n-1}\) \(-x_{n-1} )+\ldots \)
\(+a_0( \) \(x_{n-1}\) \( -x_n ) ( \)
\(x_{n-1}\) \(-x_{n-1} )\)
or \(y_{n-1}=a_n+a_{n-1}( x_{n-1}-x_n )+0+...\)
or \(y_{n-1}=a_n+a_{n-1}( -h)\)
or \(y_{n-1}=y_n-a_{n-1}h\)
or \(a_{n-1}=\frac{y_n-y_{n-1}}{h}\)
or \(a_{n-1}=\frac{\nabla {y_n}}{h}\)
Putting, \( x=x_{n-2}\), we get
\( y=a_n+a_{n-1}( x-x_n )+a_{n-2}( x-x_n )( x-x_{n-1} )+\ldots +a_0( x-x_n )\ldots ( x-x_1 )\)
or \( y_{n-2}=a_n+a_{n-1}( \) \(x_{n-2}\) \(-x_n )+a_{n-2}( \) \(x_{n-2}\) \( -x_n )(\) \(x_{n-2}\) \(-x_{n-1} )\)
\(+a_{n-3}( \) \(x_{n-2}\) \( -x_n ) ( \) \(x_{n-2}\) \(-x_{n-1} )(\) \( x_{n-2}\) \( - x_{n-2})+...\)
or \(y_{n-2}=a_n+a_{n-1}( x_{n-2}-x_n )+ a_{n-2}( x_{n-2}-x_n )( x_{n-2}-x_{n-1} )+0+...\)
or \(y_{n-2}=a_n+a_{n-1}( x_{n-2}-x_n )+ a_{n-2}( x_{n-2}-x_n )( x_{n-2}-x_{n-1} )\)
or \(y_{n-2}=a_n+a_{n-1}( 2h )+ a_{n-2}(2h)(h)\)
or \(y_{n-2}=2a_n+a_{n-1}( h )+2 a_{n-2}(h^2)\)
or \(a_{n-2}=\frac{\nabla ^2y_n}{2!h^2}\)
Similarly, we get
\(a_{n-i}=\frac{\nabla^i y_n}{i!h^i}\) for \( i=0,1,2,\ldots ,n \)
Setting , \( x={x_n}+ph\) and substituting, \( a_i\) for \( i=0,1,2,\ldots ,n \) we get
\( y=a_n+a_{n-1}( x-x_n )+a_{n-2}( x-x_n )( x-x_{n-1} )+\ldots +a_0( x-x_n )\ldots ( x-x_1 )\)
or \( y=y_n+\frac{\nabla y_n}{h}[ ph ]+ \frac{\nabla ^2y_n}{2!h^2} [ p( p+1 )h^2 ]+...\)
or \( y=y_n+\nabla y_n[ p ]+ \frac{\nabla ^2y_n}{2!} [ p( p+1 )]+...\)
This is Newton’s backward difference interpolation formula and it is useful for
interpolation near the end of a set of tabular values.
Example 1
The population of a town was as given below. Estimate the population for the year 1925.
| x | 1891 | 1901 | 1911 | 1921 | 1931 |
| y | 46 | 66 | 81 | 93 | 101 |
Solution
According to given set of data values, we form difference table as below
| X | Y | 1st diff | 2nd diff | 3rd diff | 4th diff |
| 1891 | 46 | ||||
| 66-46=20 | |||||
| 1901 | 66 | 15-20=-5 | |||
| 81-66=15 | (-3)-(-5)=2 | ||||
| 1911 | 81 | 12-15=-3 | (-1)-(2)=-3 | ||
| 93-81=12 | (-4)-(-3)=-1 | ||||
| 1921 | 93 | 8-12=-4 | |||
| 101-93=8 | |||||
| 1931 | 101 |
Here,
\( h=10\) and \( x_n=1931\)
Thus, we have
\( x=x_n+ph\)
or
\( 1925=1931+p\times 10\)
or
\( p=-0.6\)
Using formula, we get
\( y=y_n+\nabla y_n[ p ]+ \frac{\nabla ^2y_n}{2!} [ p( p+1 )]+...\)
or
\( y_{1925}= 101+ 8 [ -0.6 ]+ \frac{(-4)}{2!} [ (-0.6)( 0.4 )]+\frac{(-1)}{3!} [ (-0.6)( 0.4 )(1.4)]+\frac{(-3)}{4!} [ (-0.6)( 0.4 )(1.4)(2.4)]\)
or
\( y_{1925}=101-4.8+0.48+0.056+0.1008\)
or
\( y_{1925}=96.83\)
This completes.
Example 2
In the table below, values of y are consecutive terms of a series are given. Find tenth term of the series.
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 2.7 | 6.4 | 12.5 | 21.6 | 34.3 | 51.2 | 72.9 |
Solution
According to given set of data values, the difference table is
| X | Y | 1st diff | 2nd diff | 3rd diff |
| 3 | 2.7 | |||
| 3.7 | ||||
| 4 | 6.4 | 2.4 | ||
| 6.1 | 0.6 | |||
| 5 | 12.5 | 3 | ||
| 9.1 | 0.6 | |||
| 6 | 21.6 | 3.6 | ||
| 12.7 | 0.6 | |||
| 7 | 34.3 | 4.2 | ||
| 16.9 | 0.6 | |||
| 8 | 51.2 | 4.8 | ||
| 21.7 | ||||
| 9 | 72.9 |
Here,
\( h=1\) and \({x_n}=9\)
Thus, we have
\(x={x_n}+ph\)
or
\(10=9+p\)
or
\(p=1\)
Now, tenth term of the series is
\(y=y_n+p\nabla y_n+\frac{p( p+1 )}{2!}\nabla ^2y_n+\ldots \)
or
\(y=72.9+21.7+4.8\frac2{2!}+0.6\frac{2\times 3}{3!}\)
or
\(y=100\)
Example 3
Find cubic polynomial which takes the following values.
\( y( 0 )=1,y( 1 )=0,y( 2 )=1\) and \(y( 3 )=10\)
Solution
According to given set of data values, we form difference table as below
| X | Y | 1st diff | 2nd diff | 3rd diff |
| 0 | 1 | |||
| -1 | ||||
| 1 | 0 | 2 | ||
| 1 | 6 | |||
| 2 | 1 | 8 | ||
| 9 | ||||
| 3 | 10 |
Here,
\( h=1\) and \(x_n=3\)
Thus, we have
\( x=x_n+ph\)
or
\( p=(x-3)\)
Now, cubic polynomial for given data value is
\(y=y_n+p\nabla y_n+\frac{p( p+1 )}{2!}\nabla ^2y_n+\ldots \)
or
\( y=10+(x-3)(9 )+\frac{(x-3)(x-2)}{2!}(8)+\frac{(x-3)(x-2)(x-1)}{3!}(6)\)
or
\( y=10+9x-27+4(x^2-5x+6)+(x-1)(x^2-5x+6)\)
or
\( y=10+9x-27+(4x^2-20x+24)+(x^3-6x^2+11x-6)\)
or
\( y=x^3-2x^2+1\)
This completes.
Exercise
Use Newton's Backward difference formula to compute followings.- Find the value when x=4 from the following table [Answer: 33]
x 0 1 2 3 y 1 0 1 10 - Find solution of f(11) of an equation \(x^2-x+1\) using a = 2 and b= 10, step value h = 1.
-
Estimate the number of students who obtained marks between 75 and 80
Marks 30-40 40-50 50-60 60-70 70-80 Number 31 42 51 35 31 - Find cubic polynomial which takes the following values: y( 0 )=1,y( 1 )=2, y( 2 )=1 and y( 3 )=10
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